Show that the triangle whose angles satisfy the equality \[\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2\] is right angled

Four swallows are catching a fly. At first, the swallows are at the four vertices of a tetrahedron, and the fly is in its interior. Their maximal speeds are equal. Prove that the swallows can catch the fly.

Let $A$ and $B$ be two fixed points of a given circle and $XY$ a diameter of this circle. Find the locus of the intersection points of lines $AX$ and $BY$ . ($BY$ is not a diameter of the circle). Albanian National Mathematical Olympiad 2010---12 GRADE Question 1.

$A$ and $B$ are two points in the plane $\alpha$, and line $r$ passes through points $A, B$. There are $n$ distinct points $P_1, P_2, \ldots, P_n$ in one of the half-plane divided by line $r$. Prove that there are at least $\sqrt n$ distinct values among the distances $AP_1, AP_2, \ldots, AP_n, BP_1, BP_2, \ldots, BP_n.$

Given an isosceles triangle $ ABC$ with $ AB \equal{} BC$. A point $ M$ is chosen inside $ ABC$ such that $ \angle AMC \equal{} 2\angle ABC$ . A point $ K$ lies on segment $ AM$ such that $ \angle BKM \equal{}\angle ABC$. Prove that $ BK \equal{} KM\plus{}MC$.

In the interior of a square $ABCD$ we construct the equilateral triangles $ABK, BCL, CDM, DAN.$ Prove that the midpoints of the four segments $KL, LM, MN, NK$ and the midpoints of the eight segments $AK, BK, BL, CL, CM, DM, DN, AN$ are the 12 vertices of a regular dodecagon.

Given is a triangle $ABC$.On the extensions of the side $AB$ we consider points $A_1,B_1$ such that $AB_1=BA_1$ (with $A_1$ lying closer to $B$).On the extensions of the side $BC$ we consider points $B_4,C_4$ such that $CB_4=BC_4$ (with $B_4$ lying closer to $C$).On the extensions of the side $AC$ we consider points $C_1,A_4$ such that $AC_1=CA_4$ (with $C_1$ lying closer to $A$).On the segment $A_1A_4$ we consider points $A_2,A_3$ such that $A_1A_2=A_3A_4=mA_1A_4$ where $0<m<\frac{1}{2}$.Points $B_2,B_3$ and $C_2,C_3$ are defined similarly,on the segments $B_1B_4,C_1C_4$ respectively.If $D\equiv BB_2\cap CC_2 \ , \ E\equiv AA_3\cap CC_2 \ , \ F\equiv AA_3\cap BB_3$, $\ G\equiv BB_3\cap CC_3 \ , \ H\equiv AA_2\cap CC_3$ and $I\equiv AA_2\cap BB_2$,prove that the diagonals $DG,EH,FI$ of the hexagon $DEFGHI$ are concurrent. [hide=Diagram][asy]import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.984603447540051, xmax = 21.28710511372557, ymin = -6.555010307713199, ymax = 10.006614273002825; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); draw((1.1583842866003107,4.638449718549554)--(0.,0.)--(7.,0.)--cycle, aqaqaq); /* draw figures */ draw((1.1583842866003107,4.638449718549554)--(0.,0.), uququq); draw((0.,0.)--(7.,0.), uququq); draw((7.,0.)--(1.1583842866003107,4.638449718549554), uququq); draw((1.1583842866003107,4.638449718549554)--(1.623345080409327,6.500264738079558)); draw((0.,0.)--(-0.46496079380901606,-1.8618150195300045)); draw((-3.0803965232149757,0.)--(0.,0.)); draw((7.,0.)--(10.080396523214976,0.)); draw((1.1583842866003107,4.638449718549554)--(0.007284204967787214,5.552463941947242)); draw((7.,0.)--(8.151100081632526,-0.9140142233976905)); draw((-0.46496079380901606,-1.8618150195300045)--(8.151100081632526,-0.9140142233976905)); draw((-3.0803965232149757,0.)--(0.007284204967787214,5.552463941947242)); draw((10.080396523214976,0.)--(1.623345080409327,6.500264738079558)); draw((0.,0.)--(3.7376079411107392,4.8751985535596685)); draw((-0.7646359770779035,4.164347956460432)--(7.,0.)); draw((1.1583842866003107,4.638449718549554)--(5.997084862772141,-1.150964422430769)); draw((0.,0.)--(7.966133662513563,1.6250661845198895)); draw((-2.308476341169285,1.3881159854868106)--(7.,0.)); draw((1.1583842866003107,4.638449718549554)--(1.6890544250513695,-1.624864820496926)); draw((2.0395968109217,2.660375186246903)--(2.9561195753832448,0.6030390855677443), linetype("2 2")); draw((3.4388364046369224,1.909931693481981)--(1.4816619768719694,0.8229159040072803), linetype("2 2")); draw((1.3969966570225139,1.8221911417546572)--(4.301698851378541,0.8775330211014288), linetype("2 2")); /* dots and labels */ dot((1.1583842866003107,4.638449718549554),linewidth(3.pt) + dotstyle); label("$A$", (0.6263408942608304,4.2), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.44658827292841696,0.04763072114368767), NE * labelscalefactor); dot((7.,0.),linewidth(3.pt) + dotstyle); label("$C$", (7.008893888822507,0.18518574257820614), NE * labelscalefactor); dot((1.623345080409327,6.500264738079558),linewidth(3.pt) + dotstyle); label("$B_1$", (1.7267810657369815,6.6777827542874775), NE * labelscalefactor); dot((-0.46496079380901606,-1.8618150195300045),linewidth(3.pt) + dotstyle); label("$A_1$", (-1.1068523758141076,-1.6305405403574376), NE * labelscalefactor); dot((10.080396523214976,0.),linewidth(3.pt) + dotstyle); label("$B_4$", (10.062615364668826,-0.612633381742001), NE * labelscalefactor); dot((-3.0803965232149757,0.),linewidth(3.pt) + dotstyle); label("$C_4$", (-3.3077327187664096,-0.612633381742001), NE * labelscalefactor); dot((0.007284204967787214,5.552463941947242),linewidth(3.pt) + dotstyle); label("$C_1$", (0.1036318128096586,5.714897604245849), NE * labelscalefactor); dot((8.151100081632526,-0.9140142233976905),linewidth(3.pt) + dotstyle); label("$A_4$", (8.521999124602214,-1.1903644717669786), NE * labelscalefactor); dot((-2.308476341169285,1.3881159854868106),linewidth(3.pt) + dotstyle); label("$C_3$", (-2.9776006673235647,1.7808239912186203), NE * labelscalefactor); dot((-0.7646359770779035,4.164347956460432),linewidth(3.pt) + dotstyle); label("$C_2$", (-1.1618743843879151,4.504413415622086), NE * labelscalefactor); dot((1.6890544250513695,-1.624864820496926),linewidth(3.pt) + dotstyle); label("$A_2$", (1.6167370485893664,-2.125738617521704), NE * labelscalefactor); dot((5.997084862772141,-1.150964422430769),linewidth(3.pt) + dotstyle); label("$A_3$", (6.211074764502297,-1.603029536070534), NE * labelscalefactor); dot((7.966133662513563,1.6250661845198895),linewidth(3.pt) + dotstyle); label("$B_3$", (8.081823056011753,1.7808239912186203), NE * labelscalefactor); dot((3.7376079411107392,4.8751985535596685),linewidth(3.pt) + dotstyle); label("$B_2$", (3.8451283958285725,5.027122497073257), NE * labelscalefactor); dot((2.0395968109217,2.660375186246903),linewidth(3.pt) + dotstyle); label("$D$", (1.7542920700238853,2.991308179842383), NE * labelscalefactor); dot((3.4388364046369224,1.909931693481981),linewidth(3.pt) + dotstyle); label("$E$", (3.542507348672631,2.083445038374561), NE * labelscalefactor); dot((4.301698851378541,0.8775330211014288),linewidth(3.pt) + dotstyle); label("$F$", (4.22,0.93), NE * labelscalefactor); dot((2.9561195753832448,0.6030390855677443),linewidth(3.pt) + dotstyle); label("$G$", (2.909754250073844,0.10265272971749505), NE * labelscalefactor); dot((1.4816619768719694,0.8229159040072803),linewidth(3.pt) + dotstyle); label("$H$", (0.9839839499905795,0.43278478116033936), NE * labelscalefactor); dot((1.3969966570225139,1.8221911417546572),linewidth(3.pt) + dotstyle); label("$I$", (0.9839839499905795,1.8908680083662353), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

Let $ABC$ be a triangle, and $D$, $E$, $F$ points on the segments $BC$, $CA$, and $AB$ respectively such that $$ \frac{BD}{DC} = \frac{CE}{EA} = \frac{AF}{FB}. $$ Show that if the centres of the circumscribed circles of the triangles $DEF$ and $ABC$ coincide, then $ABC$ is an equilateral triangle.

Let $\omega$ be a semicircle and $AB$ its diameter. $\omega_1$ and $\omega_2$ are two different circles, both tangent to $\omega$ and to $AB$, and $\omega_1$ is also tangent to $\omega_2$. Let $P,Q$ be the tangent points of $\omega_1$ and $\omega_2$ to $AB$ respectively, and $P$ is between $A$ and $Q$. Let $C$ be the tangent point of $\omega_1$ and $\omega$. Find $\tan\angle ACQ$.

A trapezoid with bases $AB$ and $CD$ is inscribed into a circle centered at $O$. Let $AP$ and $AQ$ be the tangents from $A$ to the circumcircle of triangle $CDO$. Prove that the circumcircle of triangle $APQ$ passes through the midpoint of $AB$.

On the sides of triangle \( ABC \), points \( D_1, D_2, E_1, E_2, F_1, F_2 \) are chosen such that when going around the triangle, the points occur in the order \( A, F_1, F_2, B, D_1, D_2, C, E_1, E_2 \). It is given that \[ AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2. \] Prove that the perimeters of the triangles formed by the lines \( AD_1, BE_1, CF_1 \) and \( AD_2, BE_2, CF_2 \) are equal.

A hexagonal table is given, as the one on the drawing, which has $~$ $2012$ $~$ columns. There are $~$ $2012$ $~$ hexagons in each of the odd columns, and there are $~$ $2013$ $~$ hexagons in each of the even columns. The number $~$ $i$ $~$ is written in each hexagon from the $~$ $i$-th column. Changing the numbers in the table is allowed in the following way: We arbitrarily select three adjacent hexagons, we rotate the numbers, and if the rotation is clockwise then the three numbers decrease by one, and if we rotate them counterclockwise the three numbers increase by one (see the drawing below). What's the maximum number of zeros that can be obtained in the table by using the above-defined steps.

Let $ABC$ be a triangle with $A',B',C'$ are midpoints of $BC,CA,AB$ respectively. The circle $(\omega_A)$ of center $A$ has a big enough radius cuts $B'C'$ at $X_1,X_2$. Define circles $(\omega_B), (\omega_C)$ with $Y_1, Y_2,Z_1,Z_2$ similarly. Suppose that these circles have the same radius, prove that $X_1,X_2, Y_1, Y_2,Z_1,Z_2$ are concyclic.

Let $ABC$ be a triangle, let $H$ be the orthocentre and $L,M,N$ the midpoints of the sides $AB, BC, CA$ respectively. Prove that \[HL^{2} + HM^{2} + HN^{2} < AL^{2} + BM^{2} + CN^{2}\] if and only if $ABC$ is acute-angled.

Cut a right triangle with an angle of $30^o$ into three isosceles non-acute triangles, among which there are no congruent ones. (Maria Rozhkova)

Let $M$ be the midpoint of the side $AC$ of a triangle $ABC$ and let $H$ be the foot of the altitude from $B$. Let $P$ and $Q$ be orthogonal projections of $A$ and $C$ on the bisector of the angle $B$. Prove that the four points $H,P,M$ and $Q$ lie on the same circle.

Let $ABC$ be an acute angled triangle. Let $B' , A'$ be points on the perpendicular bisectors of $AC, BC$ respectively such that $B'A \perp AB$ and $A'B \perp AB$. Let $P$ be a point on the segment $AB$ and $O$ the circumcenter of the triangle $ABC$. Let $D, E$ be points on $BC, AC$ respectively such that $DP \perp BO$ and $EP \perp AO$. Let $O'$ be the circumcenter of the triangle $CDE$. Prove that $B', A'$ and $O'$ are collinear. [i]Steve Dinh[/i]

Let $ABC$ be an acute triangle with orthocenter $H$, and let $P$ be a point on the nine-point circle of $ABC$. Lines $BH, CH$ meet the opposite sides $AC, AB$ at $E, F$, respectively. Suppose that the circumcircles $(EHP), (FHP)$ intersect lines $CH, BH$ a second time at $Q,R$, respectively. Show that as $P$ varies along the nine-point circle of $ABC$, the line $QR$ passes through a fixed point. [i]Proposed by Brandon Wang[/i]

Let $D$ be the midpoint of the base $BC$ of an isosceles triangle $ABC,$ $E$ be the point at the side $AC$ such that $\angle CDE=60^\circ,$ and $M$ be the midpoint of $DE.$ Prove that $\angle AME=\angle BMD.$

Let $ ABCD$ be a parellogram with $ AB>AD$. Suposse the ratio between diagonals $ AC$ and $ BD$ is $ \frac {AC} {BD}\equal{}3$. Let $ r$ be the line symmetric to $ AD$ with respect to $ AC$ and $ s$ the line symmetric to $ BC$ with respect to $ BD$. If $ r$ and $ s$ intersect at $ P$ , find the ratio $ \frac {PA} {PB}$ Daniel

Points $A(6,13)$ and $B(12,11)$ lie on circle $\omega$ in the plane. Suppose that the tangent lines to $\omega$ at $A$ and $B$ intersect at a point on the $x$-axis. What is the area of $\omega$? $\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) } \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}$

Given are three circles $(O_1), (O_2), (O_3)$, pairwise intersecting each other, that is, every single circle meets the other two circles at two distinct points. Let $(X_1)$ be the circle externally tangent to $(O_1)$ and internally tangent to the circles $(O_2), (O_3),$ circles $(X_2), (X_3)$ are defined in the same manner. Let $(Y_1)$ be the circle internally tangent to $(O_1)$ and externally tangent to the circles $(O_2), (O_3)$, the circles $(Y_2), (Y_3)$ are defined in the same way. Let $(Z_1), (Z_2)$ be two circles internally tangent to all three circles $(O_1), (O_2), (O_3)$. Prove that the four lines $X_1Y_1, X_2Y_2, X_3Y_3, Z_1Z_2$ are concurrent. Nguyễn Văn Linh

An equilateral triangle $ABC$ is given. With $BC$ as diameter, a semicircle is drawn outside the triangle. On the semicircle, points $D$ and $E$ are chosen such that the arc lengths $BD, DE$ and $EC$ are equal. Prove that the line segments $AD$ and $AE$ divide the side $BC$ into three equal parts. [img]https://1.bp.blogspot.com/-hQQV-Of96Ls/XzXCZjCledI/AAAAAAAAMV0/SwXa4mtEEm04onYbFGZiTc5NSpkoyvJLwCLcBGAsYHQ/s0/2010%2BMohr%2Bp5.png[/img]

Given is a triangle $ABC$ and a point $X$ inside its circumcircle. If $I_B, I_C$ denote the $B, C$ excenters, then prove that $XB \cdot XC <XI_B \cdot XI_C$.

$ABC$ is a right-angled triangle. A square $ABDE$ is constructed on the opposite side of the hypothenuse $AB$ from $C$. The bisector of $\angle C$ cuts $DE$ at $F$. If $AC = 1$ and $BC = 3$, compute $\frac{DF}{EF}$. (A Blinkov)